Twin Paradox

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Visit the TeacherWeb® videos to learn about the features, modifications, and customizations available on your website. Click here to learn more. Can't Be Due To Turnaround Acceleration <blockquote dir="ltr" style="margin-right: 0px;"> <p><strong><u><font size="5">The Turnaround Class</font></u></strong> </p> <p><font size="5">Some contend that Einstein was wrong about how the net proper time difference accumulates in a Twin Paradox scenario. Some contend that it's caused by the turnaround acceleration or change of frames. There are many variations on this approach. </font><font size="5">The key is getting the proponent to be specific as to how S's clock accumulates proper time for segments bounded by well-defined events. </font></p> <p><font size="5"><u><strong>Key Questions/Points For This Class of Arguments</strong></u></font></p> <p><font size="5">By asking questions of the proponent of this class of reconciliation argument, we establish that they hold or agree that:</font></p> <p><font size="5">1a) All clocks in a given inertial frame tick at the same rate - i.e., accumulate proper time at the same rate.</font></p> <p><font size="5">1b) As a corollary to 1a) above, if a clock in an inertial frame accumulates a net proper time difference versus another clock in an inertial frame, then it also accumulates the same net proper time difference versus all other clocks in that inertial frame. For example, if S's clock during the turnaround accumulates a net time difference of 100 billion ticks versus the stay-at-home twin's clock at rest in R, then it will have also have accumulated a net time difference of 100 billion ticks versus all clocks at rest in R as they all accumulate proper time at the same rate.   </font></p> <font size="5"> </font> <p><font size="5"><font size="5"><strong><u>The Numbers</u></strong></font></font></p> <p><font size="5"><font size="5">Given the base parameters for the specific scenario defined in the <a target="_blank" href="apt2.aspx">Methodology</a> page, the proponent of this class of reconciliation argument would fill in the proper time intervals for the various segments of S's round trip. The numbers below were given by a physics professor who published a paper in a prestigious physics journal explaining the net time difference in terms of the effects of changing frames of reference during the turnaround:</font></font></p> <p><font size="5"><strong><u><font size="5"> <table style="border: 1px dotted #7f7c75;" width="425" border="0" cellpadding="2" cellspacing="0"> <tbody> <tr> <td style="border: 1px dotted #7f7c75;" width="61%" height="17"><strong><u><font face="Arial" size="2"> </font></u></strong> <p align="left"><strong><u><font face="Arial" size="2">Segment on S's Worldline </font></u></strong><u><font face="Arial" size="2"></font></u><font face="Arial" size="2"></font></p> </td> <td style="border: 1px dotted #7f7c75;" width="39%" height="17"><strong><u><font face="Arial" size="2"> </font></u></strong> <p align="right"><strong><u><font face="Arial" size="2"># of Proper Time Ticks</font></u></strong><u><font face="Arial" size="2"></font></u><font face="Arial" size="2"></font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" width="61%" height="17"><font face="Arial" size="2"> </font> <p align="left"><font face="Arial" size="2">E1-E1a (Initial Acceleration)</font></p> </td> <td style="border: 1px dotted #7f7c75;" width="39%" height="17"><font face="Arial" size="2"> </font> <p align="right"><font face="Arial" size="2">500</font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" width="61%" height="17"><font face="Arial" size="2"> </font> <p align="left"><font face="Arial" size="2">E1a-E2 (Constant velocity outbound)</font></p> </td> <td style="border: 1px dotted #7f7c75;" width="39%" height="17"><font face="Arial" size="2"> </font> <p align="right"><font face="Arial" size="2">(50,000,000,000 - 1,000)</font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" width="61%" height="17"><font face="Arial" size="2"> </font> <p align="left"><font face="Arial" size="2">E2-E2a (Turnaround Acceleration - Part 1)</font></p> </td> <td style="border: 1px dotted #7f7c75;" width="39%" height="17"><font face="Arial" size="2"> </font> <p align="right"><font face="Arial" size="2">500</font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" width="61%" height="17"><font face="Arial" size="2"> </font> <p align="left"><font face="Arial" size="2">E2a-E3 (Turnaround Acceleration - Part 2)</font></p> </td> <td style="border: 1px dotted #7f7c75;" width="39%" height="17"><font face="Arial" size="2"> </font> <p align="right"><font face="Arial" size="2">500</font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" width="61%" height="17"><font face="Arial" size="2"> </font> <p align="left"><font face="Arial" size="2">E3-E4a (Constant velocity inbound)</font></p> </td> <td style="border: 1px dotted #7f7c75;" width="39%" height="17"><font face="Arial" size="2"> </font> <p align="right"><font face="Arial" size="2">(50,000,000,000 - 1,000)</font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" width="61%" height="17"><font face="Arial" size="2"> </font> <p align="left"><font face="Arial" size="2">E4a-E4 (Final Deceleration)</font></p> </td> <td style="border: 1px dotted #7f7c75;" width="39%" height="17"><font face="Arial" size="2"> </font> <p align="right"><font face="Arial" size="2">500</font></p> </td> </tr> </tbody> </table> </font></u></strong></font></p> <p><font size="5"><font size="5"> <table style="border: 1px dotted #7f7c75;" width="425" border="0" cellpadding="2" cellspacing="0"> <tbody> <tr> <td style="border: 1px dotted #7f7c75;" width="61%" height="17"><strong><u><font face="Arial" size="2"> </font></u></strong> <p align="left"><strong><u><font face="Arial" size="2">R's Worldline </font></u></strong><u><font face="Arial" size="2"></font></u><font face="Arial" size="2"></font></p> </td> <td style="border: 1px dotted #7f7c75;" width="39%" height="17"><strong><u><font face="Arial" size="2"> </font></u></strong> <p align="right"><strong><u><font face="Arial" size="2"># of Proper Time Ticks</font></u></strong><u><font face="Arial" size="2"></font></u><font face="Arial" size="2"></font></p> </td> </tr> <tr> <td style="border: 1px dotted #7f7c75;" width="61%" height="17"><font face="Arial" size="2"> </font> <p align="left"><font face="Arial" size="2">E1-E4 (Whole scenario)</font></p> </td> <td style="border: 1px dotted #7f7c75;" width="39%" height="17"><font face="Arial" size="2"> </font> <p align="right"><font face="Arial" size="2">200,000,000,000</font></p> </td> </tr> </tbody> </table> </font></font></p> <p><font size="5"><font size="5">If you have looked at the <a href="apt3.aspx" target="_blank">Can't Be Due To Relative Velocity</a> page, you'll notice that the same numbers were filled in here as for that class of reconciliation argument. In fact, if one just quickly looked at the table of numbers above, one might be tempted to think along the lines "<em>The professor has attributed just 100 billion ticks for the constant velocity segments and the constant velocity segments can be made to be 99.99999...9% of the whole trip - So why bother discussing the turnaround acceleration? - It's already clear how and 'where' the net proper time difference accumulated.</em>" </font></font></p> <p><font size="5"><font size="5">However, t</font>he professor held that while the twins are both at rest in their respective inertial frames, their clocks tick in unison (i.e., their clocks accumulated proper time at the same rate). However, during the turnaround acceleration, while only 1,000 ticks of proper time accumulated for S, the claim was that 100 billion ticks accumulated for the stay-at-home twin R. This is claimed to be due, in effect, to relative simultaneity as S changes frames during the turnaround.</font></p> <p><font size="5"><font size="5"><strong><u>The Problems</u></strong></font></font></p> <font size="5"><font size="5"> </font></font> <p><font size="5"><font size="5">Some claim that it is the change in relative simultaneity that causes this huge jump in net proper time difference. However, relative simultaneity is a measure of differences in observations - differences in what different observers consider to be simultaneous - not about how proper time accumulates. One can use relative simultaneity formulae to compute the "correct numerical answer" because those formulae employ distance of separation between S and R. However, the "correct numerical answer" is in terms of observed time not proper time. Also, as we seen, many mutually exclusive solutions to the Twin Paradox have been manufactured to produce the "correct numerical answer" so  that's not sufficient to know one has the right physics answer. </font></font></p> <p><font size="5"><font size="5">Even if one held that the net proper time difference was due to the period of acceleration because of, say, some virtual gravitational field, one would still encounter the problems discussed below.</font></font></p> <p><font size="5"><font size="5">Let's see what problems arise when we try to apply this approach in general:</font></font></p> <font size="5"><font size="5"><strong> </strong></font></font> <p><font size="5"><font size="5"><strong>Problem 1)</strong> Let’s say that when S comes to rest in R at the "midpoint" of his turnaround acceleration, he comes to rest in the R-frame next to R2. Since R and R2 are at rest in R and clocks at rest in the same inertial frame accumulate proper time at the same rate, this presents some problems. </font></font></p> <dir><font size="5"><font size="5"><strong> </strong></font></font> <p><font size="5"><font size="5"><strong>Qualitative:</strong> Since both S and R2 "judge/compute" their own and the other’s proper time to accumulate two thousand or less ticks between E2 and E3, it’s hard to see how S can "really" lose 100 billion <u>proper time</u> ticks between E2 and E3 versus R2. Yet, R and R2 are accumulating proper time at the same rate.</font></font></p> <font size="5"><font size="5"><strong> </strong></font></font> <p><font size="5"><font size="5"><strong>Quantitative:</strong> <u>If one uses the same methodology for computing difference in proper time between E2 and E3 for S versus R2 as was used for S versus R</u>, one will compute a very small net proper time difference for S versus R2 because the distance separation factor is very small. In fact, the calculation yields about a 1,000 ticks net proper time difference due to the turnaround acceleration. Yet, R and R2 are accumulating proper time at the same rate. So if we apply the same logic for how much proper time difference accumulated between S and R and how much accumulated between S and R2, we get answers that differ by a factor of 50,000, yet the R and the R2 clocks are ticking in unison at rest in the R-frame. Hence, there's  a contradiction. </font></font></p> </dir><font size="5"><font size="5"><strong> </strong></font></font> <p><font size="5"><font size="5"><strong>Problem 2)</strong> The above should be sufficient. Further, we know from experience that giving multiple arguments can lead the proponent to think, for example, "<em>I can't address problem 1, however, I don't find argument 4, or Footnote 2, convincing, therefore, I'll dismiss it all.</em>" Despite this point, some other problems with this class of reconciliation argument might be thought provoking for some and even help show what's fundamentally wrong with this class of reconciliation argument and point the reader to the correct solution. So here goes.</font></font></p> <p><font size="5"><font size="5">The reconciliation argument claims that S loses 100 billion ticks in his turnaround acceleration E2-E2a-E3 versus R and, hence, versus all observers in R whose clocks are ticking in unison. In contrast, the argument says that S loses just a thousand ticks in his <u>start</u> and <u>return</u> accelerations E1-E1a and E4a-E4 vs R because the distance of separation factor is small at the start and end of the round trip. Hence, from "Key Point 1b)" above, this reconciliation argument says that S loses about 1,000 ticks in his <u>start</u> and <u>return</u> accelerations, E1-E1a and E4a-E4 respectively, versus R AND, hence, versus all observers in R whose clocks are ticking in unison.</font></font></p> <p><font size="5"><font size="5">Yet, if we have S continually going back and forth in a series of Twin Paradox scenarios, then most acceleration segments become both turnaround acceleration segments for one scenario and start/return acceleration segments for another scenario. Hence, the <u>very same acceleration segment</u> is claimed to cause very large net proper time difference between S and the clocks at rest in R and conversely to not cause a very large net proper time difference between S and the clocks at rest in R. This is a contradiction.</font></font></p> <font size="5"><font size="5"><strong> </strong></font></font> <p><font size="5"><font size="5"><strong>Problem 3)</strong> The reconciliation argument methodology implies that half the net time difference accumulates between E2 and E2a, and the other half between E2a and E3. Hence, 50 billion ticks of net time difference are claimed to accumulate between E2 and E2a and another 50 billion between E2a and E3. However, let’s say S decides to begin decelerating immediately <strong>AFTER</strong> E3 and lands at rest in R not far from R2 . Consistently using the reconciliation argument's methodology would indicate that since it was a very short trip from R2, only a very small net proper time difference would accumulate versus R2 and, hence, versus R. On the one hand, the reconciliation argument says that the total net proper time difference has accumulated when E3 happens, but on the other hand the size of the net proper time difference due to the turnaround acceleration depends on what happens AFTER E3 (i.e., how long the return leg is for an asymmetrical scenario).</font></font></p> <p><font size="5"><font size="5"><strong>Problem 4)</strong> If you have a Twin Paradox scenario <u>without</u> any turnaround acceleration, theory still predicts the same net proper time difference accumulation. See the diagram below:</font></font></p> </blockquote><font size="5"><font size="5"><font size="4"> </font></font></font> <p> </p> <pre><font size="5"><font size="5"><font size="4">t<sub>R <p><span style="font-size: large; ">\|\</span></p> </sub></font></font></font> <pre> <p><span style="font-size: large; ">\| \</span></p> <p><font size="5"><font size="5"><font size="4">\| \</font></font></font></p> <p><font size="5"><font size="5"><font size="4">\| \</font></font></font></p> <p><font size="5"><font size="5"><font size="4">\|R \S2 R2</font></font></font></p> <p><font size="5"><font size="5"><font size="4">\| /S1</font></font></font></p> <p><font size="5"><font size="5"><font size="4">\| /</font></font></font></p> <p><font size="5"><font size="5"><font size="4">\| /</font></font></font></p> <p><font size="5"><font size="5"><font size="4">\| /</font></font></font></p> <p><font size="5"><font size="5"><font size="4">\|/_______________x<sub>R</sub></font></font></font></p> </pre> </pre> <p><span style="font-size: 15px;"><br /> </span></p> <font size="5"><font size="5"><font size="4"><sub></sub></font> </font></font><blockquote dir="ltr" style="margin-right: 0px;"> <p><font size="5"><font size="5">We modify our basic scenario so that <strong>S1</strong>, instead of decelerating to come to rest with <strong>R2</strong>, simply has a near miss with <strong>R2</strong> and keeps on going. We further suppose that that near miss is really a 3-way near miss and includes <strong>S2</strong> passing by <strong>R2</strong> going "west" (left). Let us call this arbitrarily 3-way near miss "Event 2/3"as it replaces the turnaround acceleration between E2 and E3 of the old scenario. We will call <strong>S1</strong>'s initial near miss with <strong>R</strong> "Event 1" and <strong>S2</strong>'s final near miss with <strong>R</strong> "Event 4". If one computes the difference between <strong>R</strong>'s proper time between Event 1 and Event 4 and the sum of <strong>S1</strong>'s proper time between Event 1 and Event 2/3 <strong>plus</strong> <strong>S2</strong>'s proper time between Event 2/3 and Event 4, it is exactly the same as the net time difference for the Twin Paradox. This suggests that the <u>effects of the turnaround acceleration</u> are not the cause of the net proper time difference.<br /> <br /> </font></font><strong style="font-size: x-large; ">Problem 5)</strong><span style="font-size: x-large; "> Whatever </span><span style="font-size: x-large; ">time dilation factor the proponent is employing, it should be dependent on the physical parameters that define S’s turnaround acceleration. The problem is that one can define any number of round trips where S has </span><u style="font-size: x-large; ">exactly</u><span style="font-size: x-large; "> the same parameters for the acceleration (same location, and start and exit velocity, same rate of acceleration, etc., etc.) and yet this alleged explanation dictates that the same physical parameters will yield wildly varying predictions for the net proper time difference between S's clocks and the clocks at rest in R. Why should the physical effects of the very same acceleration be dependent on whether S stopped for tea at R or much further away at "R9" at some very distant point in the past?<br /> <br /> <strong>Problem 6)</strong> The data on time dilation does not match this type of claim. The data says that the time differences are a function of velocity and NOT acceleration per se or frame change per se. </span></p> <p><font size="5"><font size="5"><strong>Footnote 1 - Point to reflect on): </strong><font size="5">Suppose that at the mid-point of the scenario for R (i.e., at the 100 billion tick mark), R does the same initial start acceleration as S did, then immediately does the same turnaround acceleration and then immediately does the same final acceleration to come back to rest in R. Theory and physicists we've discussed this with agree that for this modified scenario, the net proper time difference would be approximately the same as for original scenario. Also, if we just look at this added, very short round trip by R, it would result in a relatively small amount of proper time lost versus the clocks that remained at rest in R and hence, should have a relatively small effect on the NPTD versus S when S returns. However, we see that R does the same amount of accelerating and same amount of changing of frames of reference and at the same distance of separation from S. Hence, if we consistently apply the same equations as was used to (erroneously) compute the net proper time lost by S versus R during S’s turnaround acceleration to R’s newly added turnaround acceleration, we (equally erroneously) compute the same, very large, net proper time lost by R versus S.</font>   </font></font></p> <p><font size="5"><font size="5"><strong>Footnote 2 - Point to reflect on)</strong>:  R and S start at rest next to each other with their clocks ticking in unison (i.e., accumulating proper time at the same rate.) The proponent says that the difference in net proper time is caused by the turnaround acceleration or associated change in frames. That means that during the constant velocity parts of the trip their clocks continue ticking in unison (i.e., accumulating proper time at the same rate). The difference in net proper time accumulates between E2 and E3 – during the turnaround acceleration. However, between E2 and E3, S only accumulates 1000 ticks of proper time, yet it's alleged that S accumulates 100,000,000,000 <strong>less</strong> ticks of proper time than R between E2 and E3. So unless one claims that S's clock goes backward that means that R's clock accumulates approximately 100,000,000,000 more ticks of proper time during the acceleration period where S accumulates just 1000 ticks because between all other event pairs they are claimed to be accumulating proper time at the same rate. Hence, the twins go from accumulating proper time at the same rate to having S accumulating proper time at 1/100,000,000<sup>th</sup> the rate of R. According to the proponent's math, this huge difference in clock rate is due to the huge distance in separation between S and R at the turnaround. The math  might come out OK, but the implied physics doesn't seem to make sense.</font></font></p> <p><font size="5"><br /> </font></p> </blockquote><font size="5"><font size="5"></font></font> <p><font size="3">© 2004, 2012 </font></p> Automatically clean input to optimize for the web. Or, would you like to view and optionally restore one of these previous versions? Password: Forgot your password?

Twin Paradox

Can't Be Due To Turnaround Acceleration

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